Question

A body is dropped on ground from a height 'h₁' and after hitting the ground, it rebounds to a height 'h₂'. If the ratio of velocities of the body just before and after hitting ground is 4, then percentage loss in kinetic energy of the body is 4. The value of x is

Answer 1

375

Answer 2

Let $v_1$ be the velocity just before hitting the ground and $v_2$ be the velocity just after hitting the ground. Given $\frac{v_1}{v_2} = 4$. The kinetic energy before collision is $KE_1 = \frac{1}{2}mv_1^2$ and after collision is $KE_2 = \frac{1}{2}mv_2^2$. The ratio of kinetic energies is $\frac{KE_2}{KE_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$. The percentage loss in kinetic energy is $\left(1 - \frac{KE_2}{KE_1}\right) \times 100\% = \left(1 - \frac{1}{16}\right) \times 100\% = \frac{15}{16} \times 100\% = \frac{375}{4}\%$. The question states the percentage loss is $\frac{x}{4}\%$. Equating $\frac{x}{4}\% = \frac{375}{4}\%$, we get $x=375$.